Question

Evaluate : $\underset{x\to 2}{lim}\frac{x^3+\sqrt{x+2}-10}{x^2-4}$

Answer 1

We have, $\underset{x\to 2}{lim}\frac{x^3+\sqrt{x+2}-10}{x^2-4}$

Let $L=\underset{x\to 2}{lim}\frac{x^3+\sqrt{x+2}-10}{x^2-4}$

This is $\frac{0}{0}$ form.

So, apply L-hospital rule,

$L=\underset{x\to 2}{lim}\frac{3x^2+\frac{1}{2\sqrt{x+2}}-0}{2x-0}$

$=\underset{x\to 2}{lim}\frac{3x^2+\frac{1}{2\sqrt{x+2}}}{2x}$

$=\frac{3(2{)}^2+\frac{1}{2\sqrt{2+2}}}{2\times 2}$

$=\frac{12+\frac{1}{2\sqrt{4}}}{4}$

$=\frac{12+\frac{1}{4}}{4}$

$=\frac{48+1}{16}$

$L=\frac{49}{16}$

Hence, this is the answer.